CIVE1400: An Introduction to Fluid Mechanics Unit 3: Fluid … · 2017-11-10 · 2.Demonstrate...

Unit 3: Fluid Dynamics

CIVE1400: Fluid Mechanics www.efm.leeds.ac.uk/CIVE/FluidLevel1 Lecture 8 98

CIVE1400: An Introduction to Fluid Mechanics


Dr P A Sleigh: [email protected] Dr CJ Noakes: [email protected]

January 2008

Module web site: www.efm.leeds.ac.uk/CIVE/FluidsLevel1

Unit 1: Fluid Mechanics Basics 3 lectures

Flow Pressure

Properties of Fluids Fluids vs. Solids

Viscosity

Unit 2: Statics 3 lectures Hydrostatic pressure Manometry / Pressure measurement Hydrostatic forces on submerged surfaces Unit 3: Dynamics 7 lectures The continuity equation. The Bernoulli Equation. Application of Bernoulli equation. The momentum equation. Application of momentum equation. Unit 4: Effect of the boundary on flow 4 lectures Laminar and turbulent flow Boundary layer theory An Intro to Dimensional analysis Similarity



Fluid Dynamics

Objectives

1.Identify differences between: • steady/unsteady • uniform/non-uniform • compressible/incompressible flow

2.Demonstrate streamlines and stream tubes

3.Introduce the Continuity principle

4.Derive the Bernoulli (energy) equation

5.Use the continuity equations to predict pressure

and velocity in flowing fluids

6.Introduce the momentum equation for a fluid

7.Demonstrate use of the momentum equation to predict forces induced by flowing fluids



Fluid dynamics:

The analysis of fluid in motion

Fluid motion can be predicted in the same way as the motion of solids

By use of the fundamental laws of physics and the

physical properties of the fluid

Some fluid flow is very complex: e.g.

• _____________________ • _____________________ • _____________________ • _____________________

All can be analysed

with varying degrees of success (in some cases hardly at all!).

There are many common situations

which analysis gives very accurate predictions



Flow Classification Fluid flow may be

classified under the following headings

_______________: Flow conditions (velocity, pressure, cross-section or

depth) are the same at every point in the fluid. ________________:

Flow conditions are not the same at every point.

________________: Flow conditions may differ from point to point but

DO NOT change with time.

________________: Flow conditions change with time at any point.

Fluid flowing under normal circumstances - a river for example -

conditions vary from point to point we have non-uniform flow.

If the conditions at one point vary as time passes

then we have unsteady flow.



Combining these four gives.

______________________. Conditions do not change with position

in the stream or with time. E.g. flow of water in a pipe of constant diameter at

constant velocity.

_________________________ Conditions change from point to point in the stream but

do not change with time. E.g. Flow in a tapering pipe with constant velocity at the

inlet.

_________________________ At a given instant in time the conditions at every point are

the same, but will change with time. E.g. A pipe of constant diameter connected to a pump pumping at a constant rate which is then switched off.

__________________________

Every condition of the flow may change from point to point and with time at every point.

E.g. Waves in a channel.

This course is restricted to Steady uniform flow - the most simple of the four.



Compressible or Incompressible Flow?

All fluids are compressible - even water. Density will change as pressure changes.

Under ___________ conditions

- provided that changes in pressure are small - we usually say the fluid is incompressible

- it has _____________ density.

Three-dimensional flow In general fluid flow is three-dimensional.

Pressures and velocities change in all directions.

In many cases the greatest changes only occur in

two directions or even only in one.

Changes in the other direction can be effectively ignored making analysis much more simple.



One dimensional flow:

Conditions vary only _______________________ not across the cross-section.

The flow may be unsteady with the parameters varying in time but not across the cross-section.

E.g. Flow in a pipe.

But: Since flow must be zero at the pipe wall

- yet non-zero in the centre - there is a difference of parameters across the

cross-section.

Pipe Ideal flow Real flow

Should this be treated as two-dimensional flow? Possibly - but it is only necessary if very high

accuracy is required.



Two-dimensional flow

Conditions vary in the direction of flow and in ___________________ at right angles to this.

Flow patterns in two-dimensional flow can be shown

by curved lines on a plane.

Below shows flow pattern over a weir.

��

In this course we will be considering: • ____________

• _______________ • ___________________________



Streamlines

It is useful to visualise the flow pattern. Lines joining points of equal velocity - velocity

contours - can be drawn.

These lines are know as __________________. Here are 2-D streamlines around a cross-section of

an aircraft wing shaped body:

��

Fluid flowing past a solid boundary does not flow into or out of the solid surface.

Very close to a boundary wall the flow direction

must be along the boundary.



Some points about streamlines:

• Close to a solid boundary, streamlines are ______________ to that boundary

• The direction of the streamline is the ________ of

the fluid velocity

• Fluid can not _______ a streamline

• Streamlines can not cross ______________

• Any particles starting on one streamline will stay on that same streamline

• In __________ flow streamlines can change

position with time

• In _______ flow, the position of streamlines does not change.



Streamtubes

A circle of points in a flowing fluid each has a streamline passing through it.

These streamlines make a tube-like shape known

as a streamtube

��

��

In a two-dimensional flow the streamtube is flat (in the plane of the paper):



Some points about streamtubes

• The “walls” of a streamtube are ___________

• Fluid cannot flow across a streamline, so fluid _______ _______ a streamtube “wall”.

• A streamtube is not like a pipe.

Its “walls” move with the fluid.

• In __________ flow streamtubes can change position with time

• In ________ flow, the position of streamtubes

does not change.



Flow rate

Mass flow rate

m dmdt

= = masstime taken to accumulate this mass

Volume flow rate - Discharge.

More commonly we use volume flow rate Also know as discharge.

The symbol normally used for discharge is Q.

discharge, Q volume of fluidtime

=



Discharge and mean velocity

Cross sectional area of a pipe is A Mean velocity is um.

mAu Q =

We usually drop the “m” and imply mean velocity.

Continuity Mass entering = Mass leaving + Increase per unit time per unit time of mass in control vol per unit time For steady flow there is no increase in the mass within the control volume, so For steady flow

Mass entering = Mass leaving per unit time per unit time

Q1 = Q2 = A1u1 = A2u2

Control volume

Mass flow in

Mass flow out



Applying to a streamtube:

Mass enters and leaves only through the two ends (it cannot cross the streamtube wall).

ρ1

u1

A1

ρ2 u2A2

Mass entering = Mass leaving per unit time per unit time

ρ δ ρ δ1 2A u A u1 1 2 2=

Or for steady flow,

ρ δ1 A u1 1 = = =

This is the continuity equation.



In a real pipe (or any other vessel) we use the mean velocity and write

ρ1A um1 1 = = =

For incompressible, fluid ρ1 = ρ2 = ρ (dropping the m subscript)

= =

This is the continuity equation most often used.

This equation is a very powerful tool. It will be used repeatedly throughout the rest of this

course.



Some example applications of Continuity

1. What is the outflow?

2. What is the inflow?



3. Water flows in a circular pipe which increases in diameter from 400mm at point A to 500mm at point B. Then pipe then splits into two branches of diameters 0.3m and 0.2m discharging at C and D respectively. If the velocity at A is 1.0m/s and at D is 0.8m/s, what are the discharges at C and D and the velocities at B and C?



Lecture 9: The Bernoulli Equation


The Bernoulli equation is a statement of the principle of conservation of energy along a

streamline

It can be written:

pg

ug

z1 12

12ρ+ + = H = Constant

These terms represent:

Pressure energy perunit weight

Kineticenergy perunit weight

Potentialenergy perunit weight

Totalenergy perunit weight

+ + =

These term all have units of length, they are often referred to as the following:

pressure head = velocity head =

potential head = total head =



Restrictions in application of Bernoulli’s equation:

• Flow is _________

• Density is __________ (incompressible)

• ____________ losses are __________

• It relates the states at two points along a single

streamline, (not conditions on two different streamlines)

All these conditions are impossible to satisfy at any

instant in time!

Fortunately, for many real situations where the conditions are approximately satisfied, the equation

gives very good results.



The derivation of Bernoulli’s Equation:

A

BB’

A’mg

z

Cross sectional area a

An element of fluid, as that in the figure above, has potential energy due to its height z above a datum and kinetic energy

due to its velocity u. If the element has weight mg then potential energy = mgz

potential energy per unit weight = z

kinetic energy = 12

2mu

kinetic energy per unit weight = u

g

2

2

At any cross-section the pressure generates a force, the fluid will flow, moving the cross-section, so work will be done. If the pressure at cross section AB is p and the area of the cross-section is a then force on AB = pa

when the mass mg of fluid has passed AB, cross-section AB will have moved to A’B’

volume passing AB = mg

gm

ρ ρ=

therefore



distance AA’ = maρ

work done = force × distance AA’

= pa ma

pm× =ρ ρ

work done per unit weight = pgρ

This term is know as the pressure energy of the flowing stream. Summing all of these energy terms gives

Pressure energy perunit weight

Kineticenergy perunit weight

Potentialenergy perunit weight

Totalenergy perunit weight

+ + =

or

pg

ug

z Hρ

+ + =2

2

By the principle of conservation of energy, the total energy in the system does not change, thus the total head does not change. So the Bernoulli equation can be written

pg

ug

z Hρ

+ + = =2

2Constant



The Bernoulli equation is applied along _______________

like that joining points 1 and 2 below.

1

2

total head at 1 = total head at 2

or

pg

ug

z pg

ug

z1 12

12 2

222 2ρ ρ

+ + = + +

This equation assumes no energy losses (e.g. from friction) or energy gains (e.g. from a pump) along the streamline. It can be

expanded to include these simply, by adding the appropriate energy terms:

Totalenergy per

unit weight at 1

Totalenergy per unit

weight at 2

Loss per unitweight

Work doneper unitweight

Energysupplied

per unit weight= + + −

pg

ug

z pg

ug

z h w q1 12

12 2

222 2ρ ρ

+ + = + + + + −



Practical use of the Bernoulli Equation

The Bernoulli equation is often combined with the continuity equation to find velocities and pressures

at points in the flow connected by a streamline. Example: Finding pressures and velocities within a contracting and expanding pipe.

u1

p1

u2

p2

section 1 section 2 A fluid, density ρ = 960 kg/m3 is flowing steadily through the above tube. The section diameters are d1=100mm and d2=80mm. The gauge pressure at 1 is p1=200kN/m2

The velocity at 1 is u1=5m/s. The tube is horizontal (z1=z2)

What is the gauge pressure at section 2?



Apply the Bernoulli equation along a streamline joining section 1 with section 2.

pg

ug

z1 12

12ρ+ + = + +

p p2 1= +

Use the continuity equation to find u2

A u

u

m s

1 1

2

=

= =

= /

So pressure at section 2 p

N m

kN m

22

2

=

=

=

/

/

Note how

the velocity has increased the pressure has decreased



We have used both the Bernoulli equation and the Continuity principle together to solve the problem.

Use of this combination is very common. We will be seeing this again frequently throughout the rest of

the course.

Applications of the Bernoulli Equation

The Bernoulli equation is applicable to many situations not just the pipe flow.

Here we will see its application to flow

measurement from tanks, within pipes as well as in open channels.



Applications of Bernoulli: Flow from Tanks Flow Through A Small Orifice

Flow from a tank through a hole in the side.

��

1

2

��

��

Aactual

Vena contractor

h

The edges of the hole are sharp to minimise frictional losses by

minimising the contact between the hole and the liquid.

The streamlines at the orifice contract reducing the area of flow.

This contraction is called the ______ ____________

The amount of contraction must

be known to calculate the ________



Apply Bernoulli along the streamline joining point 1 on the surface to point 2 at the centre of the orifice.

At the surface velocity is negligible (u1 = 0) and the pressure

atmospheric (p1 = 0).

At the orifice the jet is open to the air so again the pressure is atmospheric (p2 = 0).

If we take the datum line through the orifice

then z1 = h and z2 =0, leaving

h

u

=

=2

This theoretical value of velocity is an overestimate as

friction losses have not been taken into account.

A coefficient of velocity is used to correct the theoretical velocity,

uactual =

Each orifice has its own coefficient of velocity, they usually lie in the range( 0.97 - 0.99)



The discharge through the orifice

is jet area × jet velocity

The area of the jet is the area of the vena contracta not

the area of the orifice.

We use a coefficient of contraction to get the area of the jet

Aactual =

Giving discharge through the orifice:

Q Au

Q A uC C A u

C A u

actual actual actual

c v orifice theoretical

d orifice theoretical

===

=

=

Cd is the coefficient of discharge,

Cd = Cc × Cv



Time for the tank to empty We have an expression for the discharge from the tank

Q C A ghd o= 2

We can use this to calculate how long it will take for level in the to fall

As the tank empties the level of water falls.

The discharge will also drop.

��

h1h2

The tank has a cross sectional area of A.

In a time δt the level falls by δh The flow out of the tank is

Q Au

Q A

=

= −

(-ve sign as δh is falling)



This Q is the same as the flow out of the orifice so

=

= −δ δt AC A g

hhd o 2

Integrating between the initial level, h1, and final level, h2,

gives the time it takes to fall this height

[ ]

[ ]

tA

C A g

hh

C A g

C A g

d o hh

d o

d o

=−

=

=

∫2

2

2

12δ



Submerged Orifice What if the tank is feeding into another?

��

h1h2

��

Area A1

Area A2

Orifice area Ao Apply Bernoulli from point 1 on the surface of the deeper

tank to point 2 at the centre of the orifice,

pg

ug

z pg

ug

z

h

u

1 12

12 2

22

1

2

2 2

0 0

ρ ρ+ + = + +

+ + = + +

=

And the discharge is given by Q C A u

C Ad o

d o

==

So the discharge of the jet through the submerged orifice

depends on the difference in head across the orifice.



Lecture 10: Flow Measurement Devices


Pitot Tube The Pitot tube is a simple ________ ________ device.

Uniform velocity flow hitting a solid blunt body, has streamlines similar to this:

��

��

21

Some move to the left and some to the right. The centre one hits the blunt body and stops.

At this point (2) velocity is ______

The fluid does not move at this one point.

This point is known as the ____________ point.



Using the Bernoulli equation we can calculate the pressure at this point.

Along the central streamline at 1: velocity u1 , pressure p1 At the stagnation point (2): u2 = 0. (Also z1 = z2)

p u

p

1 12

2

2ρ+ =

=

How can we use this?

The blunt body does not have to be a solid.

It could be a static column of fluid.

Two piezometers, one as normal and one as a Pitot tube within the pipe can be used as shown below to measure

velocity of flow.



��

��

�

��

��

��

h2h1

1 2

We have the equation for p2 , p

gh

u

2

2

=

=

=

ρ

We now have an expression for velocity from two pressure measurements and the application of the

Bernoulli equation.



Pitot Static Tube The necessity of two piezometers makes this

arrangement awkward.

The Pitot static tube combines the tubes and they can then be easily connected to a manometer.

2

1

A B

h

X

��

1

[Note: the diagram of the Pitot tube is not to scale. In reality its diameter

is very small and can be ignored i.e. points 1 and 2 are considered to be at the same level]

The holes on the side connect to one side of a manometer, while the central hole connects to the other

side of the manometer



Using the theory of the manometer, p

p

p

p gX

A

B

A

=

=

=

+ =2 ρ

We know that p p u2 1 121

2= + ρ , giving

( )p hg

u

man1

1

+ − = +

=

ρ ρ

The Pitot/Pitot-static is:

• Simple to use (and analyse)

• Gives velocities (not discharge)

• May block easily as the holes are small.



Pitot-Static Tube Example

A pitot-static tube is used to measure the air flow at the centre of a 400mm diameter building ventilation duct. If the height measured on the attached manometer is 10 mm and the density of the manometer fluid is 1000 kg/m3, determine the volume flow rate in the duct. Assume that the density of air is 1.2 kg/m3.



Venturi Meter

The Venturi meter is a device for measuring _____________ in a pipe.

It is a rapidly converging section which ________ the velocity of flow and hence __________ the pressure.

It then returns to the original dimensions of the pipe by a

gently diverging ‘diffuser’ section.

about 6°

about 20°

��

1

2

z1

z2

datum

h



Apply Bernoulli along the streamline from point 1 to point 2

pg

ug

z pg

ug

z1 12

12 2

222 2ρ ρ

+ + = + +

By continuity Q u A u A

u u AA

= =

=

1 1 2 2

21 1

2

Substituting and rearranging gives

p pg

z z

u

1 21 2

1

−+ − = ⎛

⎝⎜⎞⎠⎟−⎡

⎣⎢⎤⎦⎥

= ⎡⎣⎢

⎤⎦⎥

=

ρ



The theoretical (ideal) discharge is u×A.

Actual discharge takes into account the losses due to friction, we include a coefficient of discharge (Cd ≈0.9)

Q u AQ C Q C u A

Q C A Ag p p

gz z

A A

ideal

actual d ideal d

actual d

== =

=

−+ −

⎡

⎣⎢⎤

⎦⎥

−

1 1

1 1

1 2

1 21 2

12

22

2ρ

In terms of the manometer readings

p gz p gh g z hp p

gz z

man1 1 2 2

1 21 2

+ = + + −

−+ − = ⎛

⎝⎜⎞⎠⎟

ρ ρ ρ

ρ

( )

Giving

Q C A Aactual d= 1 2

This expression does not include any elevation terms. (z1 or z2)

When used with a manometer

The Venturimeter can be used without knowing its angle.



Venturimeter design:

• The diffuser assures a gradual and steady _____________ after the throat. So that ________ rises to something near that before the meter.

• The angle of the diffuser is usually between ___ and ____ degrees.

• Wider and the flow might separate from the walls increasing energy loss.

• If the angle is less the meter becomes very long and pressure losses again become significant.

• The efficiency of the diffuser of increasing pressure back to the original is rarely greater than ______%.

• Care must be taken when connecting the manometer so that no burrs are present.



Venturimeter Example

A venturimeter is used to measure the flow of water in a 150 mm diameter pipe. The throat diameter of the venturimeter is 60 mm and the discharge coefficient is 0.9. If the pressure difference measured by a manometer is 10 cm mercury, what is the average velocity in the pipe? Assume water has a density of 1000 kg/m3 and mercury has a relative density of 13.6.



Lecture 11: Notches and Weirs


• A _______ is an opening in the side of a tank or reservoir.

• It is a device for measuring ___________.

• A ____ is a notch on a larger scale - usually found in rivers.

• It is used as both a discharge measuring device and a device to raise water levels.

• There are many different designs of weir.

• We will look at sharp crested weirs.

Weir Assumptions • velocity of the fluid approaching the weir is _____ so we

can ignore ________ _________. • The velocity in the flow depends only on the _____ below the

free surface. u gh= 2

These assumptions are fine for tanks with notches or reservoirs

with weirs, in rivers with high velocity approaching the weir is substantial the kinetic energy must be taken into account



A General Weir Equation

Consider a horizontal strip of width b, depth h below the free surface

��

��

b

δhH

h

velocity through the strip, discharge through the strip,

uQ Au== =δ

Integrating from the free surface, h=0, to the weir crest,

h=H, gives the total theoretical discharge

Qtheoretical =

This is different for every differently shaped weir or notch.

We need an expression relating the width of flow across

the weir to the depth below the free surface.



Rectangular Weir

The width does not change with depth so

b B= =constant

��

B

H

Substituting this into the general weir equation gives

Q B g h dhH

theoretical =

=

∫2 3 2

0

/

To get the actual discharge we introduce a coefficient of discharge, Cd, to account for

losses at the edges of the weir and contractions in the area of flow,

Q Cdactual =



Rectangular Weir Example

Water enters the Millwood flood storage area via a rectangular weir when the river height exceeds the weir crest. For design purposes a flow rate of 162 litres/s over the weir can be assumed 1. Assuming a height over the crest of 20cm and

Cd=0.2, what is the necessary width, B, of the weir? 2. What will be the velocity over the weir at this

design?



‘V’ Notch Weir The relationship between width and depth is dependent

on the angle of the “V”.

bH

θ

h

The width, b, a depth h from the free surface is

( )b H h= − ⎛⎝⎜

⎞⎠⎟

22

tan θ

So the discharge is

Q g

g

g

H

H

theoretical tan2

tan2

tan2

= ⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟ −⎡⎣⎢

⎤⎦⎥

= ⎛⎝⎜

⎞⎠⎟

∫2 2

2 2

815

2

0

0

θ

θ

θ

The actual discharge is obtained by introducing a coefficient of discharge

Q C g Hdactual tan2

= ⎛⎝⎜

⎞⎠⎟

815

2 5 2θ /



‘V’ Notch Weir Example Water is flowing over a 90o ‘V’ Notch weir into a tank with a cross-sectional area of 0.6m2. After 30s the depth of the water in the tank is 1.5m. If the discharge coefficient for the weir is 0.8, what is the height of the water above the weir?



Lecture 12: The Momentum Equation


We have all seen moving fluids exerting forces.

• The lift force on an aircraft is exerted by the air

moving over the wing.

• A jet of water from a hose exerts a force on whatever it hits.

The analysis of motion is as in solid mechanics: by

use of Newton’s laws of motion.

The Momentum equation is a statement of _________ _____ ______

It relates the sum of the forces

to the acceleration or rate of change of momentum.



From solid mechanics you will recognise F = ma

What mass of moving fluid we should use?

We use a different form of the equation.

Consider a streamtube:

And assume________ _____________ flow

��

A1u1

ρ1

A2

u2

ρ2

u1 δt



In time δt a volume of the fluid moves from the inlet a distance u1δt, so

volume entering the stream tube = area distance

= ×

The mass entering, mass entering stream tube = volume density

=×

And momentum momentum entering stream tube = mass velocity

=×

Similarly, at the exit, we get the expression:

momentum leaving stream tube =



By Newton’s 2nd Law.

Force = rate of change of momentum

F= A u t u A u t ut

( )ρ δ ρ δδ

2 2 2 2 1 1 1 1−

We know from continuity that

Q A u A u= =1 1 2 2

And if we have a fluid of constant density, i.e. ρ ρ ρ1 2= = , then

F =



An alternative derivation From conservation of mass

mass into face 1 = mass out of face 2

we can write

rate of change of mass = =

= =

mdmdt

A u A uρ ρ1 1 1 2 2 2

The rate at which momentum enters face 1 is

ρ1 1 1 1 1A u u mu=

The rate at which momentum leaves face 2 is ρ2 2 2 2 2A u u mu=

Thus the rate at which momentum changes across

the stream tube is ρ ρ2 2 2 2 1 1 1 1 2 1A u u A u u mu mu− = −

So

Force = rate of change of momentumF m u u= −( )2 1



So we have these two expressions,

either one is known as the momentum equation

F m u u

F Q u u

= −

= −

( )

( )

2 1

2 1ρ

The Momentum equation.

This force acts on the fluid in the direction of the flow of the fluid.



The previous analysis assumed the inlet and outlet velocities in the same direction i.e. a one dimensional system.

What happens when this is not the case?

θ2

θ1

u1

u2

We consider the forces by ____________ in the directions of the co-ordinate axes.

The force in the x-direction

( )

( )

F m u u

F Q u u

x

x

= −

=

= −

=

cos cos

cos cos

2 2 1 1

2 2 1 1

θ θ

ρ θ θ

or



And the force in the y-direction

( )

( )

F m u u

F Q u u

y

y

= −

=

= −

=

sin sin

sin sin

2 2 1 1

2 2 1 1

θ θ

ρ θ θ

or

The resultant force can be found by combining

these components Fy

Fx

FResultant

φ

Fresultant =

And the angle of this force

φ = ⎛⎝⎜

⎞⎠⎟



In summary we can say:

Total force rate of change of on the fluid = momentum through

the control volume

F

F

=

=or

Remember that we are working with vectors so F is in the direction of the ____________.



This force is made up of three components: FR = Force exerted on the fluid by any solid body

touching the control volume

FB = Force exerted on the fluid body (e.g. gravity)

FP = Force exerted on the fluid by fluid pressure outside the control volume

So we say that the total force, FT,

is given by the sum of these forces:

FT =

The force exerted

by the fluid on the solid body

touching the control volume is opposite to FR.

So the reaction force, R, is given by

R =



Application of the Momentum Equation

Forces on a Bend

Consider a converging or diverging pipe bend lying in the vertical or horizontal plane

turning through an angle of θ.

Here is a diagram of a diverging pipe bend.

u1

u2

A1

A2

p1

p2

45°

y

x

1m



Why do we want to know the forces here?

As the fluid changes direction a force will act ___ ___ ______.

This force can be very large in the case of water

supply pipes. The bend must be held in place to prevent _________ at the ______.

We need to know how much force a support (thrust block) must withstand.

Step in Analysis:

1.Draw a control volume 2.Decide on co-ordinate axis system 3.Calculate the total force 4.Calculate the pressure force 5.Calculate the body force 6.Calculate the resultant force



An Example of Forces on a Bend The outlet pipe from a pump is a bend of 45° rising in the vertical plane (i.e. and internal angle of 135°). The bend is 150mm diameter at its inlet and 300mm diameter at its outlet. The pipe axis at the inlet is horizontal and at the outlet it is 1m higher. By neglecting friction, calculate the force and its direction if the inlet pressure is 100kN/m2 and the flow of water through the pipe is 0.3m3/s. The volume of the pipe is 0.075m3. [13.95kN at 67° 39’ to the horizontal]

1&2 Draw the control volume and the axis system

u1

u2

A1

A2

p1

p2

45°

y

x

1m

p1 = 100 kN/m2, Q = 0.3 m3/s θ = 45° d1 = d2 = A1 = A2 =



3 Calculate the total force in the x direction

( )F Q u uT x x x= −

=

ρ 2 1

by continuity A u A u Q1 1 2 2= = , so

( )u

u

1 2

2

0 3015 4

0 30 0707

= =

= =

.. /

..

π

( )FT x = ×

=

1000 0 3.

and in the y-direction

( )( )

( )

F Q u u

QT y y y= −

=

= ×

=

ρ

ρ

2 1

1000 0 3.



4 Calculate the pressure force. F

F

F

P

P x

P y

== =

= =

= =

pressure force at 1 - pressure force at 2( 1θ θ θ0 2, )

We know pressure at the _______ but not at the ________.

we can use __________

to calculate this unknown pressure.

pg

ug

z1 12

12ρ+ + = + + +

where hf is the friction loss In the question it says this can be _______ ____

The height of the pipe at the outlet

is 1m above the inlet. Taking the inlet level as the datum:



z1 = z2 =

So the Bernoulli equation becomes:

1000001000 9 81

16 98

2 9 810

1000 9 81

4 24

2 9 8110

22

2

2

×+

×+ =

×+

×+

=.

.

. .

.

..p

p

F

F

Px

P y

= × − ×

= =

= − ×

=

100000 0 0177 2253614 45 0 0707

2253614 45 0 0707

. . cos .

. sin .

5 Calculate the body force The only body force is the force due to gravity. That is the weight acting in the -ve y direction.

FB y =

==

There are no body forces in the x direction, FB x = 0



6 Calculate the resultant force

F F F FF F F F

T x R x P x B x

T y R y P y B y

= + +

= + +

F

F

R x

R y

=

= − +=

=

= + +=

4193 6 9496 37

899 44 11266 37 735 75

. .

. . .

And the resultant force on the fluid is given by

FRy

FRx

FResultant

φ

FR =

= +=

5302 7 12901562 2. .



And the direction of application is

φ = ⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟

=

−

−

tan

tan

1

1

The force on the bend is the same magnitude but in the opposite direction

R = =

Lecture 13: Design Study 2

See Separate Handout



Lecture 14: Momentum Equation Examples


Impact of a Jet on a Plane

A jet hitting a flat plate (a plane) at an angle of 90°

We want to find the reaction force of the plate. i.e. the force the plate will have to apply to stay in

the same position.

1 & 2 Control volume and Co-ordinate axis are shown in the figure below.

��

��

u1

u2

u2

y

x



3 Calculate the total force In the x-direction

( )F Q u uT x x x= −

=

ρ 2 1

The system is symmetrical the forces in the y-direction cancel.

FT y =

4 Calculate the pressure force.

The pressures at both the inlet and the outlets to the control volume are atmospheric.

The pressure force is zero F FPx P y= = 0

5 Calculate the body force

As the control volume is small we can ignore the body force due to gravity.

F FBx B y= = 0



6 Calculate the resultant force

F F F FF F

T x R x P x B x

R x T x

= + +

= − −

=

0 0

Exerted on the fluid. The force on the plane is the same magnitude but in

the opposite direction R FR x= −

If the plane were at an angle

the analysis is the same. But it is usually most convenient to choose the axis

system ________ to the plate.

��

��

u1

u2

u3

y

x

θ



Force on a curved vane

This case is similar to that of a pipe, but the analysis is simpler.

Pressures at ends are equal at _______________

Both the cross-section and velocities

(in the direction of flow) remain constant.

θ

u1

u2y

x

��

��

1 & 2 Control volume and Co-ordinate axis are shown in the figure above.



3 Calculate the total force in the x direction

FT x =

by continuity u u QA1 2= = , so

FT x =

and in the y-direction

( )F Q uT y = −

=

ρ θ2 0sin

4 Calculate the pressure force. The pressure at both the inlet and the outlets to the

control volume is atmospheric.

F FPx P y= = 0



5 Calculate the body force No body forces in the x-direction, FB x = 0.

In the y-direction the body force acting is the weight

of the fluid. If V is the volume of the fluid on the vane then,

F gVB x = ρ

(This is often small as the jet volume is small and

sometimes ignored in analysis.) 6 Calculate the resultant force

( )

F F F F

F F

T x R x P x B x

R x T x

= + +

= =

F F F F

F F

T y R y P y B y

R y T y

= + +

= =



And the resultant force on the fluid is given by

F F FR R x R y= −2 2

And the direction of application is

φ =⎛

⎝⎜

⎞

⎠⎟−tan 1 F

FR y

R x

exerted on the fluid. The force on the vane is the same magnitude but in the opposite direction

R FR= −



SUMMARY

The Momentum equation is a statement of Newton’s Second Law

For a fluid of constant density,

Total force rate of change of on the fluid = momentum through

the control volume

( ) ( )F = =

This force acts ____ ____ _____

in the direction of the ________ of fluid.

This is the total force FT where: FT =

FR = _______ force on the fluid from any solid body

touching the control volume FB = ______ force on the fluid body (e.g. gravity) FP = ________ force on the fluid by fluid pressure



outside the control volume

We work with components of the force:

θ2

θ1

u1

u2

( ) ( )F Q u u Qx x x= − =ρ ρ2 1

( ) ( )F Q u u Qy y y= − =ρ ρ2 1

The resultant force can be found by combining

these components Fy

Fx

FResultant

φ

And the angle this force acts:

φ = ⎛⎝⎜

⎞⎠⎟

−tan 1

Fresultant = +



Lecture 15: Calculations Unit 3: Fluid Dynamics

1. The figure below shows a smooth curved vane attached to a rigid foundation. The jet of water, rectangular in section, 75mm wide and 25mm thick, strike the vane with a velocity of 25m/s. Calculate the vertical and horizontal components of the force exerted on the vane and indicate in which direction these components act.

45°

25°



2. A 600mm diameter pipeline carries water under a head of 30m with a velocity of 3m/s. This water main is fitted with a horizontal bend which turns the axis of the pipeline through 75° (i.e. the internal angle at the bend is 105°). Calculate the resultant force on the bend and its angle to the horizontal.



3. A 75mm diameter jet of water having a velocity of 25m/s strikes a flat plate, the normal of which is inclined at 30° to the jet. Find the force normal to the surface of the plate.



4. In an experiment a jet of water of diameter 20mm is fired

vertically upwards at a sprung target that deflects the water at an angle of 120° to the horizontal in all directions. If a 500g mass placed on the target balances the force of the jet, was is the discharge of the jet in litres/s?

5. Water is being fired at 10 m/s from a hose of 50mm diameter into the atmosphere. The water leaves the hose through a nozzle with a diameter of 30mm at its exit. Find the pressure just upstream of the nozzle and the force on the nozzle.

CIVE1400: An Introduction to Fluid Mechanics Unit 3: Fluid … · 2017-11-10 · 2.Demonstrate...

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Transcript of CIVE1400: An Introduction to Fluid Mechanics Unit 3: Fluid … · 2017-11-10 · 2.Demonstrate...